b To solve the inequality we must first determine its boundary point(s). To do that, we will treat the inequality as if it was an equation and solve for

x .

Inequality : Equality : 2 x^{2} - 5 x < 12 2 x^{2} - 5 x = 12

Let's solve it by using the Quadratic Formula.

2 x^{2} - 5 x = 12

SubEqn

$LHS - 12 = RHS - 12$

2 x^{2} - 5 x - 12 = 0

UseQuadForm

Use the Quadratic Formula: $a = 2, b = - 5, c = - 12$

x = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( 2 ) ( - 1 2 )}{2 ( 2 )}

▼

Evaluate right-hand side

NegNeg

$- (- a) = a$

x = \frac{5 \pm ( - 5 ) ^{2} - 4 ( 2 ) ( - 1 2 )}{2 ( 2 )}

CalcPowProd

Calculate power and product

x = \frac{5 \pm 2 5 + 9 6}{4}

AddTerms

Add terms

x = \frac{5 \pm 1 2 1}{4}

CalcRoot

Calculate root

x = \frac{5 \pm 1 1}{4}

StateSol

State solutions

x = (5 - 11) / 4 x = (5 + 11) / 4 (I) (II)

$(I), (II):$ Add and subtract terms

x = - 6 / 4 x = 16 / 4

CalcQuot

Calculate quotient

x_{1} = - 1.5 x_{2} = 4

The boundaries of the inequality is

x = - 1.5

and

= 4 .

Let's mark these solutions on a number line. We will keep the points open since this is a strict inequality. To identify where we should shade the number line, we will also include a few test points.

By substituting

x

with

- 3

0,

and

5

in the inequality we can determine where the inequality holds true.

x - 3 05 2 x^{2} - 5 x < 12 2 (- 3)^{2} - 5 (- 3) < ? 12 2 (0)^{2} - 5 (0) < ? 12 2 (5)^{2} - 5 (5) < ? 12 Evaluate 33 ≮ 12 \times 0 < 12 ✓ 25 ≮ 12 \times

As we can see, the inequality is true for values between the boundary points. Therefore, we should shade this region.

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